are you smarter than the average good third worlder student? no trigonometry allowed
>>17071195Who cares about third worlds anymore man? I am not even smarter than a clanker.
I've been thinking a lot about this class of problems a lot recently:http://lambdaplusjs35padjaiz4jw2fugdoeutse262phqr72uf634s2wdbqd.onion/math/284408So I can solve it with just elementary geometry.First, reflect B across the perpendicular bisector of AC; call the image D. BD is parallel to AC, and triangle DCP is isosceles. It is easy to work out the angles in triangles DCP and BDP. (If x is a rational number of degrees, which is most likely the case given the "no trigonometry" rule, this reduces the problem to an adventitious quadrangle problem, for which known solution methods exist.)
>>17071220Remove the parts of the diagram we no longer need.
>>17071221Basically the approach to solving these things goes like this. There's one line segment whose direction we need to find. We'll build a bunch of isosceles triangles which suffice to construct that line segment, then use them to build a sequence of congruent line segments connecting the endpoints of the line segment.You should read through Tatham's excellent writeup to understand the motivation for all this.https://www.chiark.greenend.org.uk/~sgtatham/quasiblog/adventitious/Basically, these are calculations in a convenient subfield of the complex numbers disguised as elementary geometry. Although once you know how to do it, you don't need to touch the complex numbers.In this case, it turns out that if we place point F along BD so that triangle FDP is isosceles, then triangle FBP is also isosceles. Nice.
>>17071224We're trying to connect BC with a sequence of congruent line segments, and conveniently, we can fit a triangle congruent to FBP into angle PCD as shown.
>>17071226Now erase more shit we don't need anymore.
>>17071229Now we can connect BE to form an equilateral triangle. Some easy angle chasing gives us x=30.
>>17071220>this reduces the problem to an adventitious quadrangle problem, for which known solution methods existHere's an explanation of the general method if you want to read about it:https://www.gensu.co.jp/saito/challenge/pdf/3circumcenter_d20180609.pdfI would recommend readinghttps://www.chiark.greenend.org.uk/~sgtatham/quasiblog/adventitious/first.There's also a Wikipedia article on adventitious quadrangles:https://en.wikipedia.org/wiki/Langley%27s_Adventitious_Angles
Wtf it's just 40 deg y'all fucking retarded and failed middle school math xD
>>17071195By using the sine theorem twice:sin(x) * sin(80°) = sin(80° - x) * sin(40°)x = 30°
>>17071257Nice. And pointing out 80 was twice 40 in that triangle made me realize we can do this.
>>17071237thats new to mein third worlds like mine, our entire mathematical educational background are soviet books. never heard of langley triangle, adventitious angles and adventitious quadrangle, maybe thats only an anglo thing. here its just the 80 20 triangles problems and known quadrangle set of problemsi wonder if someone training for IMO came to the same 3 circumcenter method but didnt know about its relevancy lolpage 50 is the official solution to the problemhttps://www.scribd.com/document/608037497/GEOMETRIA-OLIMPIADAS-MATEMATICA-IIIkind of similar to yourshere is the same problem but with different angles and different construction to solve ithttps://www.youtube.com/watch?v=Jsz76Jubv88
>>17071416The problem in the video is not the same problem and is actually much easier, pic related.I don't know of any school that teaches about adventitious quadrangles. Langley's Adventitious Angles is a popular problem that has been posted on /sci/ and other math-related imageboards before. It is "adventitious" in the sense that the angle to be found happens to have a rational number of degrees, but this is only the case for particular values of the given angles. Give the same problem with different numbers, and the answer is often irrational. It makes one wonder, what's going on with these particular numbers that make the solution rational? So some mathematicians studied which rational values of the given angles in the problem make the solution also rational. They called these cases "adventitious quadrangles." You can find links to papers classifying all the adventitious quadrangles in the Wikipedia article.I should clarify that I didn't use the three circumcenter method to solve the problem. The three circumcenter method can solve any adventitious quadrangle problem, but sometimes there's an easier way as there was in this case.The law of sines is another potential way to solve these. Even if you add a rule like "no trigonometry", if you have a way of getting the answer with the law of sines and proving it exactly correct using trig identities, you can rewrite the solution after the fact so that trigonometric functions are not explicitly mentioned. This is what I would recommend a high school student do if they asked for help on such a problem.
>>17071416>>17071564>The problem in the video is not the same problem and is actually much easier, pic related.Oh, on second reading, I see you said "with different angles," so what you said was correct. The particular values of the angles in these problems can change the nature and difficulty of the solutions a lot.
>>1707119530°
>>17071636[math]\pm 0.1^\circ[/math]
>>17071195APB = 80 degreesBPC = 100 degreesAB = PCBPA:AB::CBP:PCBPA/2=BAPBPC/2=CBPCBP=50 degreesBCP= 180 degrees - BPC (100 degrees) - CBP (50 degrees)BCP = 30 degrees
>>17072897yeah well fuck you too buddy
>>17072897doesn't make any sense
>>17073378Which part is confusing you, little bro?
>>17073859For starters the angle APB is 280°.
>>17072897>BPA:AB::CBP:PCAB = PC but BPA is not CBP
